Hello can someone help me?

[granite arch]

here

[jade wraith]

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[hexed beacon]

Sounds interesting, cause I think it's false

For example, for $n=2$ \
$\frac{n^2}{n^2+1} = \frac{4}{5}$ \
and $a_2 = \frac{4}{5} + \frac{4}{6} = \frac{22}{15} > \frac{4}{5}$. \
So $\forall n \in \mathbb{N}^*, a_n <= \frac{n^2}{n^2+1}$ is false

[tender echo]

Yojda

[thin wind]

Yeah the left inequality seems true though

[hexed beacon]

Yes

Hint : $a_n = n^2 * \left( \frac{1}{n^2 + 1} + \frac{1}{n^2 + 2} + ... + \frac{1}{n^2 + n} \right)$

[tender echo]

Yojda

[thin wind]

You could also say k+n^2<=n+n^2 thus n^2/(k+n^2)>=n^2/(n+n^2)

[hexed beacon]

But $a_n$ is the sum

[tender echo]

Yojda

[hexed beacon]

Wait

No it seems correct

Then by sum, a_n >= n^2/(n^2 + n)

[thin wind]

Yea but when you sum you’ll get n^3/(n+n^2)>=n^2/(n^2+n)

[thin wind]

Yes

[hexed beacon]

No because $a_n = \sum_{k=1}^n \frac{n^2}{n^2 + k}$. \
But, we know that $\forall k \in {1,2,...,n}, \frac{n^2}{n^2 + k} >= \frac{n^2}{n^2 + n}$. \
Then $\forall n \in \mathbb{N}^, a_n >= n \frac{n^2}{n^2 + n} >= \frac{n^2}{n^2 + n}$

[tender echo]

Yojda

[thin wind]

That’s what I meant

[hexed beacon]

Okay I maybe did not understand what you said but I think I've said the same thing x)

[thin wind]

It’s my bad I wrote confusingly

[hexed beacon]

np

[granite arch]

hello,so the right inequality is false ?

[hexed beacon]

yes

[granite arch]

does that mean i cant solve the problem?